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    <p>Here is a version in python (versions 2 and 3):</p> <pre><code>def aggregate(data): data.sort() i = 0 while i &lt; len(data) - 1: while i &lt; len(data) - 1 and data[i][1] &gt;= data[i+1][0]: data[i] = (data[i][0], max(data[i][1], data[i+1][1])) data.pop(i+1) i += 1 if __name__ == '__main__': itervals = [(1,2), (5,7), (9,12), (3,4), (5,6), (13,17)] formatted = lambda vals: '[{}]'.format(', '.join('({}-{})'.format( iterval[0], iterval[1]) for iterval in sorted(vals))) print(formatted(itervals)) aggregate(itervals) print(formatted(itervals)) </code></pre> <p>Output:</p> <pre class="lang-none prettyprint-override"><code>[(1-2), (3-4), (5-6), (5-7), (9-12), (13-17)] [(1-2), (3-4), (5-7), (9-12), (13-17)] </code></pre> <p>Note: this mutates a list of tuples in-place. A slightly more generic one, that will work with a list of iterables can be done by changing the line:</p> <pre><code>data[i] = type(data[i])((data[i][0], max(data[i][1], data[i+1][1]))) </code></pre> <p>Alternatively, if you know you have a list of mutable iterables, you could use:</p> <pre><code>data[i][1] = max(data[i][1], data[i+1][1]) </code></pre> <p>An alternative version, that may be a little easier to follow, is:</p> <pre><code>def aggregate(data): if not data: return data inputs = sorted(data) result = [inputs[0]] for next0, next1 in inputs[1:]: last0, last1 = result[-1] if next0 &lt;= last1: result[-1][1] = max(next1, last1) else: result.append([next0, next1]) return result </code></pre> <p>Note that this one is designed for a list of lists.</p>
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    1. POhow to overlap intervals efficiently
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    1. USMatthew Schinckel
    1. USMatthew Schinckel
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    1. COWouldn't it be a little better to replace the outer `while` with a `for i in xrange(len(data)):` and lose the `i += 1` at the end? Also the first portion of the inner `while`'s conditional, the `i < len(data) - 1` part, seems redundant.
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      1. USmartineau
    2. CONot quite. Because the inner loop contains a ``.pop()`` on the list, it needs to re-check that it won't be exceeding the bounds of the list next time it tries to access ``i+1``. Because this function mutates the original data structure, ``xrange(len(data))`` will not work (it is evaluated at the first iteration, but may end up with fewer iterations.
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      1. USMatthew Schinckel
    3. COHaving said that, I'm happy to see a rewrite of this with tests:) [Note: I am using code very similar to this in production].
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      1. USMatthew Schinckel
 

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